# Question Video: Finding Unknowns in Matrix Equations Mathematics • 10th Grade

Given that 𝐴 = [3, 0, −2 and −2, 1, 𝑥 and 0, 3, 7], find 𝑥 such that 𝐴² = [9, −6, −20 and −8, −20, −52 and −6, 24, 28].

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### Video Transcript

Given that 𝐴 is the three-by-three matrix three, zero, negative two, negative two, one, 𝑥, zero, three, seven, find the value of 𝑥 such that 𝐴 squared is equal to the three-by-three matrix nine, negative six, negative 20, negative eight, negative 20, negative 52, negative six, 24, 28.

In this question, we are given a three-by-three matrix 𝐴 containing an unknown value of 𝑥. We need to find this value of 𝑥. To do this, we are given the matrix 𝐴 squared. We can recall that we define squaring a matrix in the same way we square a number. We multiply it by itself, so 𝐴 squared is 𝐴 times 𝐴. We can use this to find 𝐴 squared. We need to multiply 𝐴 by itself. This gives us the following product for 𝐴 squared. Before we calculate the square of this matrix, we can replace 𝐴 squared in the equation with the given matrix 𝐴 squared.

We now recall that we multiply matrices by finding the sum of the products of each entry in each row of the first matrix with the corresponding elements in the columns of the second matrix. For instance, we can apply this to the second row of the first matrix and the third column of the second matrix to get the element in the second row and third column of 𝐴 squared. Equating the entry in the second row and third column of 𝐴 squared with the sum of the products of these entries gives us the following linear equation in 𝑥.

We can then solve this equation for 𝑥. We simplify each side of the equation. We then subtract four from both sides of the equation to obtain negative 56 equals eight 𝑥. Then, we divide through by eight to get that 𝑥 equals negative seven.